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Bài 1:
29.(85-47)+85.(47-29)
=29.38+85.18
=1102+1530
=2632
BÀI 2
a, 15x = -75
x =( -75) : 15
x = -5
Vậy x= -5
b,-5x+8=-27
-5x = -27 - 8
-5x= -35
x =-35 : -5
x = 7
Vậy x = 7
c,3.(17 + x ) = - 42
17 + x = - 42 : 3
17 + x = -14
x = - 14 - 17
x = -31
Vậy x = - 31
d,-10 -( 2x -16) = 0
2x - 16 = 10-0
2x - 16 = 10
2x = 10 +16
2x = 26
x = 26 : 2
x = 13
Vậy x = 13
e,( mik ko bít làm)
f,( tự làm nghen bạn)
Bài 1
a, 32.(-64) - 64. 68
= -64.(32 + 68)
= -64 .100
= - 6400
b, -54.76 + 12.(-76) - 76.34
= -76.(54 + 12 + 34)
= -76.100
= -7600
19- 42.(-19) + 38.5
= 19 + 42.19 + 19.2.5
= 19.(1 + 42 + 10)
= 19.53
= 1007
a) x- 18 = - 25
x =( -25 ) + 18
x = -7
b) 42 - x = 53 - 60
16 - x = 65
x = 16 - 65
x = -49
a: \(575-\left(2x+70\right)=445\)
=>\(2x+70=575-445=130\)
=>\(2x=130-70=60\)
=>x=60/2=30
b: \(575-2\left(x+70\right)=445\)
=>\(2\left(x+70\right)=575-445=130\)
=>x+70=130/2=65
=>x=65-70=-5
c: \(x^5=32\)
=>\(x^5=2^5\)
=>x=2
d: \(\left(3x-1\right)^3=8\)
=>\(\left(3x-1\right)^3=2^3\)
=>3x-1=2
=>3x=3
=>\(x=\dfrac{3}{3}=1\)
e: \(\left(x-2\right)^3=27\)
=>\(\left(x-2\right)^3=3^3\)
=>x-2=3
=>x=5
f: \(\left(2x-3\right)^2=9\)
=>\(\left[{}\begin{matrix}2x-3=3\\2x-3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=3\\x=0\end{matrix}\right.\)
g: \(2x+5=3^4:3^2\)
=>\(2x+5=3^2\)
=>2x+5=9
=>2x=9-5=4
=>x=4/2=2
h: \(\left(4x-5^2\right)\cdot7^3=7^4\)
=>\(4x-25=\dfrac{7^4}{7^3}=7\)
=>4x=25+7=32
=>\(x=\dfrac{32}{4}=8\)
1, \(\dfrac{x-1}{2009}+\dfrac{x-2}{2008}=\dfrac{x-3}{2007}+\dfrac{x-4}{2006}\)
\(\Leftrightarrow\left(\dfrac{x-1}{2009}-1\right)+\left(\dfrac{x-2}{2008}-1\right)=\left(\dfrac{x-3}{2007}-1\right)+\left(\dfrac{x-4}{2006}-1\right)\) ( Trừ mỗi vế cho 2 ta được phương trình như này nhé ! )
\(\Leftrightarrow\dfrac{x-2010}{2009}+\dfrac{x-2010}{2008}=\dfrac{x-2010}{2007}+\dfrac{x-2010}{2006}\)
\(\Leftrightarrow\dfrac{x-2010}{2009}+\dfrac{x-2010}{2008}-\dfrac{x-2010}{2007}-\dfrac{x-2010}{2006}=0\)
\(\Leftrightarrow\left(x-2010\right)\left(\dfrac{1}{2009}+\dfrac{1}{2008}-\dfrac{1}{2007}-\dfrac{1}{2006}\right)=0\)
Do \(\dfrac{1}{2009}+\dfrac{1}{2008}-\dfrac{1}{2007}-\dfrac{1}{2006}\ne0\) nên \(x-2010=0\Leftrightarrow x=2010\)
2, \(\dfrac{59-x}{41}+\dfrac{57-x}{43}+\dfrac{55-x}{45}+\dfrac{53-x}{47}+\dfrac{51-x}{49}=-5\)
\(\left(\dfrac{59-x}{41}+1\right)+\left(\dfrac{57-x}{43}+1\right)+\left(\dfrac{55-x}{45}+1\right)+\left(\dfrac{53-x}{47}+1\right)+\left(\dfrac{51-x}{49}+1\right)=0\)
\(\Leftrightarrow\dfrac{100-x}{41}+\dfrac{100-x}{43}+\dfrac{100-x}{45}+\dfrac{100-x}{47}+\dfrac{100-x}{49}=0\) \(\Leftrightarrow\left(100-x\right)\left(\dfrac{1}{41}+\dfrac{1}{43}+\dfrac{1}{45}+\dfrac{1}{47}+\dfrac{1}{49}\right)=0\) Do \(\dfrac{1}{41}+\dfrac{1}{43}+\dfrac{1}{45}+\dfrac{1}{47}+\dfrac{1}{49}\ne0\) nên \(100-x=0\Leftrightarrow x=100\)
5^2x-1=5^3.5^4
5^2x-1=5^7
=>2x-1=7
còn lại bn lm tiếp
Câu b hình như bn vik sai đề bài , đề bài đúng phải thế này :
7^x-3.4=296
7^x-12=296
7^x =296+12
7^x = 308
7^x = 7^3
=>x=3
c) 2^x.2^x+3=64^2:2^5
(2^x)^2 +3 = (2^6)^2 : 2^5
(2^x)^2 +3 = 2^12 : 2^5
(2^x.2) +3= 2^7